3.323 \(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=116 \[ \frac {3 i \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

3*I*a^(7/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-3*I*a^3*(a+I*a*tan(d*x+c))^(1/2)/d
-I*a^3*(a+I*a*tan(d*x+c))^(3/2)/d/(a-I*a*tan(d*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3487, 47, 50, 63, 206} \[ -\frac {i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {3 i \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((3*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((3*I)*a^3*Sqrt[a + I*a*Tan[
c + d*x]])/d - (I*a^3*(a + I*a*Tan[c + d*x])^(3/2))/(d*(a - I*a*Tan[c + d*x]))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{3/2}}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}+\frac {\left (3 i a^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+x}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}+\frac {\left (3 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}+\frac {\left (6 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {3 i \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.41, size = 137, normalized size = 1.18 \[ -\frac {i \sqrt {2} e^{-4 i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (3 e^{i (c+d x)}+e^{3 i (c+d x)}-3 \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) (a+i a \tan (c+d x))^{7/2}}{d \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(3*E^(I*(c + d*x)) + E^((3*I)*(c + d*x)) - 3*Sqr
t[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*(a + I*a*Tan[c + d*x])^(7/2))/(d*E^((4*I)*(c + d*x))*Sec[
c + d*x]^(7/2))

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fricas [B]  time = 2.47, size = 235, normalized size = 2.03 \[ -\frac {6 \, \sqrt {2} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{7}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - 6 \, \sqrt {2} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{7}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - \sqrt {2} {\left (-4 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 12 i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/4*(6*sqrt(2)*sqrt(-a^7/d^2)*d*log(4*(a^4*e^(I*d*x + I*c) + sqrt(-a^7/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) - 6*sqrt(2)*sqrt(-a^7/d^2)*d*log(4*(a^4*e^(I*d*x + I*c
) + sqrt(-a^7/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) -
sqrt(2)*(-4*I*a^3*e^(3*I*d*x + 3*I*c) - 12*I*a^3*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.26, size = 412, normalized size = 3.55 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (3 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}+3 i \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )+3 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}+3 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+8 i \left (\cos ^{4}\left (d x +c \right )\right )-4 i \left (\cos ^{3}\left (d x +c \right )\right )-8 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 i \cos \left (d x +c \right )-4 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) a^{3}}{2 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

-1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(3*I*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)+3*I*arctanh(1/
2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
3/2)*sin(d*x+c)+3*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+3*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)+8*I*cos(d*x+c)^4-4*I*cos(d*x+c)^3-8*cos(d*x+c)^3*sin(d*x+c)+4*cos(d*x+c)^
2*sin(d*x+c)-4*I*cos(d*x+c)-4*cos(d*x+c)*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)*a^3

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maxima [A]  time = 0.65, size = 117, normalized size = 1.01 \[ -\frac {i \, {\left (3 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 4 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4} - \frac {4 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5}}{i \, a \tan \left (d x + c\right ) - a}\right )}}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-1/2*I*(3*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) + 4*sqrt(I*a*tan(d*x + c) + a)*a^4 - 4*sqrt(I*a*tan(d*x + c) + a)*a^5/(I*a*tan(d*x + c) - a))/
(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

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